### Video Transcript

The speed-time graph shows the
first 55 seconds of Michael’s journey on his scooter. Michael travelled a total distance
of 650 meters in the 55 seconds. Work out the value of 𝑣.

There are two things we can work
out from this speed-time graph aside from the speed at any given point of the
journey: the first is the acceleration that’s given by the gradient of the graph at
a given point; the second is the total distance travelled, which is given by the
area under the curve, that’s the area between the graph and the 𝑥-axis.

This part of the question is asking
us to find the value of 𝑣. However, it also tells us the total
distance travelled, which means we’ll need to find the area under the graph. We can begin by splitting the shape
up into parts. We can either choose three parts or
possibly four, depending on how confident you are finding the area of a
trapezium. Let’s stick with three for now.

Between zero and 20 seconds, we
have a triangle. The formula for area of a triangle
is a half multiplied by its base multiplied by its height. The width of the base of this
triangle is 20 and its height is 𝑣. The formula then becomes a half
multiplied by 20 multiplied by 𝑣. Since a half of 20 is 10, the area
of the first part of this graph is 10𝑣.

The third shape is also a
triangle. So let’s calculate the area of the
third shape now. The width of the base of this
triangle is calculated by subtracting 40 from 55. That gives us 15. Its height is 𝑣 plus five. So the area of this triangle is a
half multiplied by 15 multiplied by 𝑣 plus five. A half of 15 is a decimal. So instead, we’ll expand the
brackets. 15 multiplied by 𝑣 is 15𝑣 and 15
multiplied by five is 75. That means that the area of the
third shape is 15𝑣 plus 75 all over two.

The second shape is a
trapezium. The formula for area of a trapezium
is a half multiplied by 𝑎 plus 𝑏 or multiplied by ℎ, where 𝑎 and 𝑏 are the
lengths of the parallel sides of the trapezium and ℎ is the height between them. The lengths of the parallel sides
of this trapezium are 𝑣 and 𝑣 plus five and the height between them is 20 that’s
found by subtracting 20 from 40.

We can simplify the brackets by
collecting like terms to give us two 𝑣 plus five. And since multiplication is
commutative, that means you can do it in any order. We can move the 20 next to the
half. A half of 20 is 10. Then, finally, we can expand the
brackets. 10 multiplied by two 𝑣 is 20𝑣 and
10 multiplied by five is 50. The area of the trapezium is 20𝑣
plus 50.

Remember we said that the total
area under the graph represents the distance travelled. In our case, that’s 650 meters. So we can add together our
expressions and make them equal to 650 to find an equation in terms of 𝑣. Now, this fraction is making our
life a little tricky. So we’ll multiply everything in
this equation by two to get rid of the two on the denominator. That gives us 1300 is equal to 20𝑣
plus 40𝑣 plus 100 plus 15𝑣 plus 75.

We can then collect together like
terms to give us 1300 is equal to 75𝑣 plus 175. Next, we’ll subtract 175 from both
sides of the equation, giving us 1125 is equal to 75𝑣. Finally, we need to divide both
sides of the equation by 75.

We can use long division to help us
perform that calculation. One 75 is 75 and that’s the most
number of 75s we can use to make 112. We then subtract 75 from 112, which
gives us 37. We bring down this remaining five
and work out how many 75s we need to make 375. Five 75s are 375. And then, once again, we subtract
375 from 375, giving us zero. That means 1125 divided by 75 is 15
with no remainder. 𝑣 is equal to 15.

Describe Michael’s acceleration for
each part of the 55-second journey.

Before we do anything, let’s change
the value of 𝑣 on our diagram. 𝑣 was 15. Therefore, 𝑣 plus five is 20. The two values on our 𝑦-axis then
representing speed are 15 and 20. Now, remember we said that the
gradient of the graph tells us the acceleration at a given point. We can therefore find the gradient
of each part of the journey using the formula gradient is change in 𝑦 over change
in 𝑥. That sometimes called rise over
run.

Let’s consider the first part of
the journey. The change in 𝑦 is the rise. It’s the height of that triangle,
which is 15. The change in 𝑥 is the run. It’s the width of the triangle,
which we worked out earlier to be 20. The gradient is therefore 15 over
20, which we can simplify to three-quarters by dividing by the numerator and the
denominator by five.

Now, let’s consider the second part
of the journey. The height of this triangle which
represents the change in 𝑦 is five. It’s the difference between 20 and
15. The width of the triangle which
represents the change in 𝑥 is 20. So the gradient is five over 20,
which once again we can simplify by dividing both the numerator and the denominator
by five. The gradient during the second part
of the journey is a quarter.

Finally, let’s consider the third
part of our journey. This part of the graph is sloping
down. That means it has a negative
gradient and represents a deceleration. The change in 𝑦 is therefore
negative 20, given by the height of the triangle. The change in 𝑥 is 15. So the gradient of this part of the
journey is negative 20 over 15. That simplifies to negative four
over three.

We mustn’t forget that when writing
the gradient as an acceleration, we need to include units — in this case that’s
metres per second squared. For the first 20 seconds, he is
accelerating at a constant rate of three-quarters metres per second squared. For the next 20 seconds, he is
accelerating at a constant rate of a quarter metres per second squared. For the final 15 seconds, he is
decelerating at a constant rate of four-thirds metres per second squared.